If F is a subfield of $\mathbb{R}$ and $\sqrt{2} +\sqrt{3} \in F$ then both $\sqrt{2}$ and $\sqrt{3}$ in F?

Question

I'm trying to show that $\sqrt{2},\sqrt{3} \in$ F given that F is a subfield of $\mathbb{R}$ and $\sqrt{2} +\sqrt{3} \in F$.

Here is my work so far

$$\sqrt{2} +\sqrt{3} \in F$$ $$(\sqrt{2} +\sqrt{3})^2 = (5+ 2\sqrt{2}\sqrt3) \in F \Longrightarrow \sqrt{2}\sqrt3 \in F$$

After this part I'm not sure what to do. I feel like both the sum and the product of $\sqrt{2}$, $\sqrt{3}$ being in the field should imply that the individual elements are in the field but I'm not if this is correct or how to justify it. I was wondering whether someone can tell me if I'm on the right track or not and give a hint as to how the proof goes.

================================================================== Would this work

$$\sqrt{2}\sqrt{3}(\sqrt{2} +\sqrt{3}) = 2\sqrt{3}+3\sqrt{2} \in F$$ $$2\sqrt{3}+3\sqrt{2} -2(\sqrt{2}+\sqrt{3}) =\sqrt{2}\in F $$ Since $\sqrt{2}$ and $\sqrt{6}$ are in F then it follows that $\sqrt{3}$ is in F just by dividing by $\sqrt{2}$

Answer 1

Since $F$ is a field and $\sqrt{2}+\sqrt{3} \in F$, its inverse is also in $F$. That is ,

$$(\sqrt{2}+\sqrt{3})^{-1}=\frac{1}{\sqrt{2}+\sqrt{3}}.\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}=-1(\sqrt{2}-\sqrt{3}) \in F $$

Hence $\sqrt{2}+\sqrt{3}+(-\sqrt{2}+\sqrt{3})=2 \sqrt{3} \in F$ and so $\sqrt{3} \in F$

Similarly $\sqrt{2} \in F$

Answer 2

Hint: Consider $(\sqrt2+\sqrt3)^3$ instead.